My personal thoughts so far is that you have to get a definition of convergence of curves straight.

Like, if you consider the plane and consider the set of not self intersecting continuos curves. Then this says that if a sequence of curves g_n converges to a curve g*, then it's possible that L(g_n) doesn't converge to L(g*), where L: C-> R is the map which maps a closed curve onto its length.

The issue then is how to define convergence, or a distance.

Perhaps by looking at the area in between both curves. (don't really know about Triangle inequality at the fly, though)

Hence, with this distance, the example shows that the function L is not continuous.
With as a consequence, you get weird things.

(it shouldn't be hard to show, given the sequence g_n here, that g_n converges to a circle for n-> infinity)

Hence,

i can let people to think about this one. for those who doesn't know the vigorous definition of riemann integral:

suppose f(x) is a continous function. divide the x-axis into arbitary small intevals[a, x_1], [x_1, x_2], [x_2, x_3],...[x_n, x_(n+1)], [x_(n+1), b], such that the maximum length of each interval is always less than an arbitary number e>0. call the partition P_e. let w_k(f) denote the max value of |f(x)-f(y)|, where x, y is in [x_k, x_(k+1)]. the value of
int(a to b)(f(x) is given by the value which the sum w_1(f)|a-x_1|+...+w_k(f)|x_k-x_(k+1)|+..
. converges when e tends to 0.

using this fact, show that what's wrong with the arc length argument. (let f(x)=integrand of the arc length formula, then divide the circle into 4 sectors. why does the value 4 is an overestimation of pi?)

At 9/3/11 10:44 AM, i-am-ghey wrote:
using this fact, show that what's wrong with the arc length argument. (let f(x)=integrand of the arc length formula, then divide the circle into 4 sectors. why does the value 4 is an overestimation of pi?)

Are you referring to the problem I stated?

yep. the method of evaluating the arc length in the problem is simply wrong.

if you repeat the process infinitely many times, you get a partition in which each interval has length less than e. you get a series of right angled triangles too. if you sum up the length of the longest side of each triangle, it will give you the real arc length. however, it is clear that the total length of two other sides of each triangle is greater than the length of the longest side.

At 9/3/11 12:03 PM, i-am-ghey wrote: yep. the method of evaluating the arc length in the problem is simply wrong.

if you repeat the process infinitely many times, you get a partition in which each interval has length less than e. you get a series of right angled triangles too. if you sum up the length of the longest side of each triangle, it will give you the real arc length. however, it is clear that the total length of two other sides of each triangle is greater than the length of the longest side.

As I see it, there really isn't an issue in that.

If you consider g_n the n-th step in the process, it's clear that the length of g_n is always equal to 4.
Now if you calculate the area in between g_n and the circle, it's certainly a decreasing sequence and I wager that it approaches 0 arbitrarily close, if you pick n large enough.
So it should still hold.

At every step, length and area of the surface enclosed by g_n can be easily calculated, using simple discrete things.

Eventually it's all about limits.

well. the triangle inequality is a strict inequality though. if the longest side of infinitesimal small triangle k is L_k, sum of the other two sides is s_k,

if you write the arc length as: (l_1)+(l_2)+...
then (l_1)+...<(s_1)+(s_2)+.... can't be equal even at the limit.

moreover, the fact that (l_1)+... does give the actual arc length comes from the defintion of a riemann integral.

just the length.

consider a straight line first. then consider a continuous function deviates slightly from the striahgt line, but the end points intersect.

then the change in arc length is very slight. however, one can also find a such a continuous functino such that the arc length is less than the sum of the of two sides of each infintesimal small traingle, which is also a portion of a circle. then it is clear that the method gives an overestimation of the arc length. but i don't think it is necessary to do so because i believe it also applies to any non-constant function.

At 9/6/11 08:15 AM, MrPercie wrote: http://www.newgrounds.com/bbs/topic/1270 356

Im going to use this [ symbol to mean square root because cant find the symbol on my keyboard

heres the questions

Q1: 3[10 X 6[10 = My answer: 28 but then I used box/grid method which got me 37[10
Real answer: 180

Q2: 2[5 X 2[5 = My answer: 13[5 real answer: 20

so was I right or wrong

I think you've got confused somwhere along the way. At some point you seem to have added where you should've multiplied. Could you show your working?

I got 180 and 20 as well.
1) 3[10*6[10 = 3*6*([10)^2 = 18*10 = 180

2) 2[5*2[5 = 4*5 = 20

At 9/3/11 02:16 PM, i-am-ghey wrote: consider a straight line first. then consider a continuous function deviates slightly from the striahgt line, but the end points intersect.

I'm sorry for my extremely late response. There's very little factual calculation to go about. At each step, the dented square has length 4 for sure, as the transformation leaves the total length invariant.

What you get is the paradox that a same area an be encased by cuves of different length (or at least approached.

The thing is quite similar to all those riddles like to prove that 1=2 by division by 0. It's just neat to finally see a more subtle side of mathematis played out.

dammit. i screwed up in one of my math tests. got a zero for the easiest question because i forgot the definitions. i was too nervous.
but i managed to complete the rest of the paper though. some of the later questions were highly non-trivial and only a small percentage of us got them.

here is a basic analysis problem: the limit superior of a sequence (denoted as limsup{x_n}is defined as the maximum real number to which a subsequence of {x_n} converges. prove that the following statements are equilavent:

statement 1: limsup(x_n)=A
statement 2: for all e_1>0, there exists natural number k_1 such that for all n>k_1, x_n<a +e_1. in addition, for every e_2>0, k_2 is integer, there exists at least one n>k_2 such that x_n>A-e_2.

i used to be best at math in my class but this year ive been sick alot and we switched math teacher from the
best iv'e ever had to an unlicensed teacher that couldn't count for shit and switched again to a pissed old man
so i almost failed the my last test also could someone give a general list of the general mathematical signs in the american (english where ever its from) and what they're called, i only know the swedish system and even though the "system" is global there are some differences and these signs are alien to me

At 11/22/11 10:48 AM, laughatyourfuneral wrote:

SOME OF THESE signs are alien to me

I'm not amazing at math, but there was a math problem going around facebook that over 100,000 people had answered wrong. Not everyone has to be great at math, but it was something that I taught to fifth graders and it completely blew my mind that full grown people couldn't figure the answer out. The question below:

40 + 40 x 0 + 1 =

At 11/22/11 10:48 AM, laughatyourfuneral wrote: i used to be best at math in my class but this year ive been sick alot and we switched math teacher from the
best iv'e ever had to an unlicensed teacher that couldn't count for shit and switched again to a pissed old man
so i almost failed the my last test also could someone give a general list of the general mathematical signs in the american (english where ever its from) and what they're called, i only know the swedish system and even though the "system" is global there are some differences and these signs are alien to me

http://en.wikipedia.org/wiki/Mathematica l_symbol
could this be what you are looking for?

seeing this thread has received replies, i guess i will try to illustrate one common technique to solve basic math problems (especially in the olympiad).

pigeonhole principle: if we place n distinct elements into m different boxes, where n>m, then at least one of the box must contain more than 1 element.

proof: trivial.

example 1: suppose 51 distinct integers are chosen from the set {1,2,...100}. there exists at least a pair of numbers which do not have common prime divisor.

proof: group it into (1,2),...(99,100). since 51 numbers are chosen, because of pigeonhole principle, two consective integers must be chosen, say k, k+1. but it is impossible that a prime p divides both numbers and p>=2.

example 2: putnam (2002). show that if 5 points are chosen arbitary from a sphere, then at least four points must lie on the same hemisphere.

proof: draw a circle connecting any of the two points. since there are only two hemispheres to contain the other three points, the result is a direct consequence of pigeonhole principle.

questions:

1. prove that the decimal representations of a rational number must be eventually periodic. (a rational number is a number that can be written in p/q, where p q are integers)

(for example: 1/4=0.2500000.....
1/7=0.142857142857...
15/26=0.57692307692307...) and so on.

hint: division algorithm.

2. consider any nine points in three dimension space, all of which have integer coordinate. prove that at least one line segment with end points selected from the nine points must contain a third point with integer coordinates.

hint: think about the condition for both endpoints.

What's after calculus?

Hello,
I have a issue with probabilities.
Let X be random variable such that EX=0, EX^2=1, EX^4<\infty. Let a>=1 and suppose that P(|X| [\in] [a,2a])=P(|X|>=a)-P(|X|>=2a)>=B.
What is the probability that P(|X|[\in] [a, \infty])=P(|X|>=a)>=?
Thanks in advance!

At 11/22/11 11:10 AM, Fro wrote:

40 + 40 x 0 + 1 =

41, right?

At 11/25/11 04:04 AM, ZJ wrote:

At 11/22/11 11:10 AM, Fro wrote:

40 + 40 x 0 + 1 =

41, right?

Yup. The multiplication takes priority, then you do the addition. These kind of questions get posted on Facebook quite a lot these days and I was surprised to see that even some of my university chums were answering them wrongly.

so that's the final answer. sinx/n=sinx=6

people can't be serious all the time, especially on the internet.

also, one small remark to question 1: sometimes a real number can have more than one decimal representation, when it has a finite decimal expansion (terminating). you can easily see this by noting 0.25=0.250000..=0.249999....

as for infinte decimal expansions, the representation must be unique. (why?)

Hey guys, I need a little help with my Maths work. I've been doing A-Level maths for less than a term so sorry if my question may seem a little amateur.

Here's the question:

---------

The function f is defined by:
f(x) = (ax + b) / (cx + d), x =/= -d / c, b =/= 0, c =/= 0.

(a) Prove that if a + d = 0, then f(x) = f^-1(x).
[NOTE: f^-1(x) being the inverse function].

(b) Prove that if a + d =/= 0 and (a - d)^2 + 4bc = 0, then the graph of y = f(x) intersects the graph of y = f^-1(x) in exactly one point.

---------

I managed to do part (a). I found the inverse of f(x) which is f^-1(x) = (b- dx) / (cx - a). Then I considered f(x) = f^-1(x) and after a bigass simplification I found that (a + d) is common in each term of the equation and the only way for the equation to be satisfied is if (a + d) is 0, so I proved that (at least I hope I did it right).

Part (b) has me stumped though. I'd love some explanation on what to do. The Maths teacher gave this question to us but he didn't expect any of us to actually manage it (in fact I was the only one in the class who even managed to do part (a)) so it's no biggie if nobody helps out, but all the same a bit of assistance will really be appreciated.

Thanks in advance.

hi. i will try to use the fact that a quadratic equation has a repeated root if and only if its discriminant is 0. (i am sure you have learnt this)

(b) f^(-1)=(b-dx)/(cx-a) aned f=(ax+b)/(cx+d)

to find the intersetion, equation the two expressions to get (b-dx)/(cx-a)=(ax+b)/(cx+d).

get rid of the denomiators by multiplication, and expand then collect the terms, yielding (ac+cd)x^2+(d^2-a^2)x-(ab+bd)=0.

the discriminant is given by (d^2-a^2)^2+4(ac+cd)(ab+bd)
factor out the expression, getting, ((a+d)^2)((d-a)^2+4bc)
but by the given condition, this is exactly zero.

so the graph only intersect at one point. done.

At 11/25/11 03:08 PM, Supersteph54 wrote:

I managed to do part (a). I found the inverse of f(x) which is f^-1(x) = (b- dx) / (cx - a). Then I considered f(x) = f^-1(x).

oops. i didn't read that part.

you shouldn't consider f equals to its inverse because that would be illogical.

since a+d=0, we have a=-d. simply replace the original formula this way:
f(x)=(ax+b)/(cx+d), =>(-dx+b)/(cx-a)=f^(-1)(x).

good job trying to increase post count.

Whoa, thanks for the help! I swear, I will never get how some people can seemingly think of this stuff so easily. I would've never guessed that I should change a + d = 0 into a = -d and interchange a and -d...

But I have a question. Why not use f(x) = f^-1(x)? I mean the way I see it, it still proves the point because if a + d =/= 0 then f(x) wouldn't be equal to its inverse. And in the question it said "Prove that if a + d = 0, then f(x) = f^-1(x).", so I thought it only makes sense to use f(x) = f^-1(x).

Thanks again, and sorry if my questions are kinda stupid. I'm terrible at this graphs stuff.

well... suppose you want to show A implies B, where A is a given condition, B is what you want to prove.

if you consider the fact that B is true, then you proceed to show that A must be true, what you have proved is that B implies A, but not the other way round.

let's say statement A is n is an even number, statement B is n equals 2.
consider n=2, if you want B to be true, A has to be true as well. but certainly A does not imply B.

if you want to show A implies B, and you want to start with B, you need to assume B is false, and try to contradict with statement A, then A implies B.

or just do a direct proof and show A implies B directly.

Sorry for the late answer.

Anyway, I see exactly what you mean. I worked it out on a piece of paper and it was correct. I'll keep this stuff in mind next time I have a bigass/challenging Maths question on functions.

...Right now I'm doing coordinate geometry, and my hatred for this topic is inconceivable. I think soon after, we'll be starting differentiation. I wonder what that's like.

differentiation can be a very easy topic if they just focus on the computational part. (for instance, using standard formula to compute the derivative of a smooth function, and evaluate the slope at a given point, sketching certain curves, checking differentiablity at a given point from defintion.)

sometimes, we are interested in finding the limit of f(x)/g(x) as x tends to a and both f(x) and g(x) either tends to 0 at a or goes to infinity. there is a computational formula called l'hopital's rule to evaluate the limit. but i think there is one point about this rule that most teachers don't mention in class, this rule is only valid if the limit f'/g' (f' and g' are derivatives) exist!!!!

but this kind of computation is straight forward, and everyone can master it.

differentiation of a single variable function can also be a relatively hard topic for high school students, if they focus on the theoritcal part (for example: using mean value theorem, inverse function theorem, rolle theorem to prove certain results. you may refer to some of the previous questions i set in this thread.) you need to have some intuition to do this kinds of proofs. but i don't think they put too much emphasis on this. so don't worry.

At 3/12/04 04:46 PM, BWS wrote: x^2+y^2=r^2

for how long has that been true!?

Quick trigonometry question.

---------

Find the general solution of the equation:
cos (x - (pi / 4)) = cos (4x + (pi / 4))

...where x is an angle in radians.

---------

The answer's supposed to be:
2(pi)(n) / 3 - (pi / 6) and
2(pi)(n) / 5

...where n is the constant used to signify how many periods the cos graph will have to make.

My answer is:
- (4(pi)(n) + pi) / 6 and 2(pi)(n) / 5

...which is very close to the result but if you open up - (4(pi)(n) + pi) / 6 you get:
- 2(pi)(n) / 3 - (pi / 6)

...and that's wrong.

I'm certain I'm doing something really stupid; some sort of numerical mistake somewhere, because I know the method I'm using is correct. If someone could shed some light I'd appreciate it a lot, but if not then it's no biggie since I can just ask the teacher on Monday.

I've included a screenshot of my work (P.V. and S.V. are the primary and secondary values respectively). It's a little careless, but meh.

actually, your solution should be correct. i see nothing wrong with both the steps and the solution. (i checked it one by one on a calculator)

although i would just simply move the other cosine expression to one side and use trig. formula to write the term into product of two trig. functions. i found it somewhat easier and more straightforward to approach the problem this way. but yeah... yours is fine.

At 1/7/12 08:53 AM, i-am-ghey wrote: actually, your solution should be correct. i see nothing wrong with both the steps and the solution. (i checked it one by one on a calculator)

although i would just simply move the other cosine expression to one side and use trig. formula to write the term into product of two trig. functions. i found it somewhat easier and more straightforward to approach the problem this way. but yeah... yours is fine.

Thanks a lot, I'm relieved to hear that.

Even in the problem after it, my answer wasn't matching with the answer book's because of a wrong sign. Perhaps it's because in the answer book, they took
4x + (pi / 4) = 2(pi)(n) + x - (pi / 4)
rather than how I took it;
x - (pi / 4) = 2(pi)(n) + 4x + (pi / 4)

Or it could be a mistake or something, I dunno.

Thanks again, much appreciated.